Question

f(x)=-x²-4x+9 qual as coordenadas do vértice da função quadrática anterioor

Answer 1

A coordenadas do vértices são dadas por (Xv, Yv)

$f(x)=-x^2-4x+9\\\\\\ Xv= \frac{-b}{2a}\\\\ Xv= \frac{-(-4)}{2*-1}\\\\ Xv= \frac{4}{-2}\\\\ Xv=-2\\\\\\ Yv=- \frac{\delta}{4a}\\\\ Yv=- \frac{(-4)^2-4*-1*9}{4*-1}\\\\ Yv=- \frac{16+36}{-4}\\\\ Yv=- \frac{52}{-4}\\\\ Yv=-(-13)\\\\ Yv=13$

As coordenadas do vértice são (-2, -13)

Answer 2

-x²
a < 0
max = yv

Δ
16 +36
52

yv = -52/-4
yv = 13

xv = 4/-2
xv = -2